Question

use a \\(\chi^2\\)-test to test the claim \\(\sigma^2 = 0.55\\) at the \\(\alpha = 0.10\\) significance level using sample statistics \\(s^2 = 0.526\\) and \\(n = 18\\). assume the population is normally distributed.

identify the null and alternative hypotheses.

a. \\(h_0: \sigma^2 = 0.55\\)
\\(h_a: \sigma^2 \
eq 0.55\\)

b. \\(h_0: \sigma^2 \geq 0.55\\)
\\(h_a: \sigma^2 < 0.55\\)

c. \\(h_0: \sigma^2 \leq 0.55\\)
\\(h_a: \sigma^2 > 0.55\\)

d. \\(h_0: \sigma^2 \
eq 0.55\\)
\\(h_a: \sigma^2 = 0.55\\)

identify the standardized test statistic.
16.26 (round to two decimal places as needed.)

identify the critical value(s).
(round to two decimal places as needed. use a comma to separate answers as needed.)

Explanation:

Step1: Determine degrees of freedom

The degrees of freedom for a chi - square test of variance is given by \(df=n - 1\). Here, \(n = 18\), so \(df=18 - 1=17\).

Step2: Determine the significance level and tails

The significance level \(\alpha = 0.10\) and since the alternative hypothesis is \(H_{a}:\sigma^{2}
eq0.55\) (two - tailed test), we split \(\alpha\) into two equal parts. So \(\alpha/2=0.05\) and \(1-\alpha/2 = 0.95\).

Step3: Find the critical values

We need to find \(\chi_{0.95}^{2}\) and \(\chi_{0.05}^{2}\) with \(df = 17\).
Using the chi - square distribution table or a statistical calculator:

• For \(\chi_{0.95}^{2}\) with \(df = 17\), we look up the value in the chi - square table. The value of \(\chi_{0.95}^{2}(17)\) is \(8.672\).
• For \(\chi_{0.05}^{2}\) with \(df = 17\), we look up the value in the chi - square table. The value of \(\chi_{0.05}^{2}(17)\) is \(27.587\).

Answer:

8.67, 27.59 (rounded to two decimal places)